I know the article of Hamilton on the inverse function theorem of Nash and Moser (with the same title) where he proves that $C^\infty(M)$ is a tame Fréchet space, when $M$ is closed or compact with boundary.

However, as far as I see no word is lost on the case of non-compact $M$. In particular, what if $M$ is an open subset of $\mathbb{R}^n$? Of course, we need to endow this with its Fréchet topology (uniform $C^m$ convergence on compact subsets).

I assume that this may not be a tame space, but I couldn't find a reference. At least it has a sequence of seminorms of increasing strength that induce the topology, by taking an exhaustion with compact sets and simultaneously letting the order of differentiation increase.

"However, as far as I see no word is lost on the case of non-compact $M$." Hamilton's proof uses the Fourier transform after embedding the compact manifold $X$ into $\mathbb{R}^d$ by Whitney. How do you do this for $M$ a unbounded open subset of $\mathbb{R}^n$?
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Willie WongAug 5 '13 at 8:47

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Also, intent behind the sentence in the final paragraph starting with "At least it has a sequence..." is true for any Fréchet space pretty much by definition.
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Willie WongAug 5 '13 at 8:48

2 Answers
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The idea behind a tame Frechet space is that you have smoothing operators with controlled estimates by which you can counteract loss of derivatives in a nonlinear operator so that a Banach fixed point argument works. You look at a formal version of this. Of course you can have smoothing operators on noncompact manifolds, but the best use for them is if you describe your space as a projective limit (intersection) of Sobolev spaces. See

for a comprehensive description of situations where you can use the Hamilton-Nash-Moser technique.

Edit: No, $C^\infty(M)$ with the compact $C^\infty$ topology is not tame.
Here by tame I mean being a direct summand in a space of rapidly
falling sequences in a Banach space.

The idea is as follows:
Just take $M=\mathbb R$ with seminorms $\|f\|_{n,k}=\sup_{|x|\le n} \max_{\alpha\le k}|f^{(\alpha)}(x)|$. If it were tame you could find constants $C_{n,k}$ such that
$C_{n,k}\|f\|_{n,k}$ is rapidly decreasing (even bounded is enough -- then you can tighten the constants). But you always have functions $f$ with $f^{(k)}(n)$ going off faster than the constants you chose.

$f\mapsto e^f$ is a counter-example to any kind of inverse function theorem on $C^\infty(\mathbb R)$.

Kofi has already accepted Peter Michor's answer, but anyway I post the following to give a different argument for seeing that $C^\infty(\Omega)$ is not tame. Tameness in the sense of Hamilton implies the interpolation inequalities $$\|x\|_{i+j}\le C_{i,j+k}\|x\|_i^{\frac k{j+k}}\|x\|_{i+j+k}^{\frac j{j+k}}$$ for a taming sequence $\nu_i:x\mapsto\|x\|_i$ of seminorms. This in turn implies existence of a continuous norm on the space. Indeed, every $\nu_i$ is necessarily a norm, since otherwise there would exist a nonzero $x$ with $\|x\|_i=0$ for all $i$, and then the space would not be Hausdorff. Consequently, for $\Omega$ a (nonempty) open set in a Euclidean space, the space $C^\infty(\Omega)$ is not tame.